On 2/25/2010 7:23 AM, prasad_chand wrote:
> Hi,
>> I use python to do simple math problems as a hobby.
>> I have made a program that finds the number of divisors(factors) of a
> given number. I am hoping to improve my language skills, specifically
> I would like to re-write the function "prime_factors" more gracefully.
> While the program works right, I am hoping that I could get some input
> on how to write better python code. I have attached the code below.
>>> def prime_factors(n):
> """
> Reduce a number to its prime factors. e.g. 48 is 2^4,3^1 (add (4+1)
> (1+1) = 10)
>> Updates a global dictionary(my_dict) with prime numbers and number
> of occurances. In the case of 48 {2:4,3:1}
>> """
> tmp_n = n
A name meaning "temporary value of n" doesn't suggest how it's being
used in the algorithm. In my implementation (see below), I used the name
"last_result", which is (a little bit) better.
>> while True:
>> if tmp_n == 1:
> break
>> tmp_check = tmp_n
>> for x in range(2,int(ceil(sqrt(tmp_n)) + 1)):
> if tmp_n % x == 0:
> add_to_dict(x)
This function changes the value of a global variable, *my_dict*. Using
global variables is frowned upon. In this case, it would be much better
to have the dictionary be local to the *prime_factor* function. After
you've broken out of the *while* loop, just execute "return my_dict".
> tmp_n /= x
> break
>> if tmp_check == tmp_n: #number is prime...so just to add to
> dict
> add_to_dict(tmp_n)
> break
The only reason that the *tmp_check* variable exists is to test whether
you fell out of the *for* loop without finding any divisors for *tmp_n*.
A cleaner approach is to use the optional *else* clause of the *for*
loop, which is executed only if you didn't *break* out of the loop:
for x in range(2,int(ceil(sqrt(tmp_n)) + 1)):
if tmp_n % x == 0:
add_to_dict(x)
tmp_n /= x
break
else:
# tmp_n is prime...so just to add to dict
add_to_dict(tmp_n)
break
>>> def add_one(x):
> return x+1
>>> def mul(x,y):
> return x * y
>> def add_to_dict(p_num):
> if my_dict.has_key(p_num):
> my_dict[p_num] += 1
> else:
> my_dict[p_num] = 1
>
As poster pruebauno pointed out, using a collections.defaultdict
eliminates the need for the *add_to_dict* function.
>> my_dict = {}
>>> prime_factors(135)
> l = map(add_one,my_dict.values())
> print reduce(mul, l, 1)
This may seem trivial, but ... don't use the single-character lowercase
"l" as a variable. It looks too much like the digit "1" -- in some
fonts, it looks identical!
FWIW, here's my implementation. It's much slower, because it doesn't use
the square root optimization. It uses another optimization: when a prime
factor is located, *all* of its occurrences are factored out at the same
time.
#--------------------------------
from collections import defaultdict
def prime_factors(n):
"""Return the prime factors of the given number (>= 2)"""
if n < 2:
print "arg must be >= 2"
return
last_result = n
factors = defaultdict(int)
next_divisor = 2
while True:
while last_result % next_divisor == 0:
factors[next_divisor] += 1
last_result /= next_divisor
if last_result == 1:
return factors
next_divisor += 1
#--------------------------------
HTH,
John